Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int toNumber(string s) {
  int Number;
  if (!(istringstream(s) >> Number)) Number = 0;
  return Number;
}
string toString(int number) {
  ostringstream ostr;
  ostr << number;
  return ostr.str();
}
long long ele(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    return ele(a * a, b / 2);
  } else {
    return a * ele(a * a, b / 2);
  }
}
long long mcd(long long a, long long b) {
  if (a == 0) return b;
  return mcd(b % a, a);
}
double d_abs(long a, long b) {
  if (a > b) {
    return a - b;
  }
  return b - a;
}
void solve() {
  int a, b, n;
  vector<long long> v;
  int k, q;
  int m;
  cin >> n >> m;
  int grid[n][m];
  int ma[n];
  set<int> rows;
  int todas = 600;
  for (int i = 0; i < (int)(n); i++) {
    int maxi = 600;
    for (int j = 0; j < (int)(m); j++) {
      cin >> grid[i][j];
      maxi = min(maxi, grid[i][j]);
    }
    ma[i] = maxi;
    for (int j = 0; j < (int)(m); j++) {
      grid[i][j] -= maxi;
    }
    todas = min(todas, maxi);
  }
  set<pair<int, int> > cols;
  bool pase[m];
  int co[m];
  int todascol = 600;
  for (int i = 0; i < (int)(m); i++) pase[i] = false;
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)(m); j++) {
      if (grid[i][j]) {
        pase[j] = true;
        set<int> se;
        for (int q = 0; q < (int)(n); q++) {
          se.insert(grid[q][j]);
        }
        if (se.size() > 1) {
          cout << -1 << endl;
          return;
        } else {
          co[j] = *se.begin();
        }
        todascol = min(todascol, *se.begin());
      } else {
        todascol = 0;
        co[j] = 0;
      }
    }
  }
  if (todas) {
    if (n > m) {
      for (int i = 0; i < (int)(n); i++) ma[i] -= todas;
      for (int j = 0; j < (int)(m); j++) co[j] += todas;
    }
  }
  if (todascol) {
    if (m > n) {
      for (int i = 0; i < (int)(n); i++) ma[i] += todascol;
      for (int j = 0; j < (int)(m); j++) co[j] -= todascol;
    }
  }
  int res = 0;
  for (int i = 0; i < (int)(n); i++) {
    res += ma[i];
  }
  for (int i = 0; i < (int)(m); i++) {
    res += co[i];
  }
  cout << res << endl;
  for (int i = 0; i < (int)(n); i++) {
    for (int can = 0; can < (int)(ma[i]); can++)
      cout << "row " << i + 1 << endl;
  }
  for (int i = 0; i < (int)(m); i++) {
    for (int can = 0; can < (int)(co[i]); can++)
      cout << "col " << i + 1 << endl;
  }
}
int main() {
  int T;
  T = 1;
  while (T--) {
    solve();
  }
}